Question 0cd0c

Jun 17, 2016

For a zero-order reaction.

Explanation:

The rate of a reaction is simply a measure of the change in the concentration of the reactants or of the products of a chemical reaction per unit of time.

$\text{rate" = "change in concentration"/"unit of time}$

For a general $n$-order reaction in which

overbrace("A" + "A" + ... + "A")^(color(blue)(n)) -> " products"

you can use the differential rate law to express the relationship between the rate of the reaction, the rate constant $k$, and the concentration of the reactant $\left[\text{A}\right]$

"rate" = ["A"]^color(blue)(n)" " " "color(orange)("(*)")

The concentration of the reactants is measured in

["A"] = "mol L"^(-1)

The rate of the reaction usually expresses the change in concentration of the reactant per second, which means that the units for the rate of the reaction are

${\text{rate" = "mol L"^(-1) "s}}^{- 1}$

If you use these units in equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$, you will get

"mol L"^(-1)"s"^(-1) = k * ("mol L"^(-1))^color(blue)(n)

Rearrange to isolate $k$ on one side of the equation

$k = {\left({\text{mol L"^(-1)"s"^(-1))/("mol L}}^{- 1}\right)}^{\textcolor{b l u e}{n}}$

You want $k$ to have the same units as the rate of the reaction

$k = {\text{mol L"^(-1)"s}}^{- 1}$

so plug this into the above equation and solve for $\textcolor{b l u e}{n}$

"mol L"^(-1)"s"^(-1) = ("mol L"^(-1)"s"^(-1))/("mol L"^(-1))^color(blue)(n)#

As you can see, the only way to have a valid equality here is if $\textcolor{b l u e}{n} = 0$, since

${\left({\text{mol L}}^{- 1}\right)}^{0} = 1$

This of course implies that you have

$\text{mol L"^(-1)"s"^(-1) = ("mol L"^(-1)"s"^(-1))/1 = "mol L"^(-1)"s"^(-1)" " " } \textcolor{g r e e n}{\sqrt{}}$

Therefore, the rate constant of a zero-order reaction has the same units as the rate of the reaction, i.e. ${\text{mol L"^(-1)"s}}^{- 1}$.